with the collaboration of Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam´ s Erd´ lyi, Zachary Franco, Christian Friesen, Ira M. Gessel, L´ szl´ a e a o Lipt´ k, Frederick W. Luttmann, Vania Mascioni, Frank B. Miles, Bogdan Petrenko, a Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ullman, Charles Vanden Eynden, Sam Vandervelde, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before July 31, 2010. Additional information, such as generalizations and references, is welcome. The problem number and the solver's name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available.

Supremum of a Nonlinear Functional 11366 [2008, 462]. Proposed by Nicolae Anghel, University of North Texas, Denton, TX. Let : R R be a continuously differentiable function such that (0) = 0 and is strictly increasing. For a &gt; 0, let Ca denote the space of all continuous funca tions from [0, a] into R, and for f Ca , let I ( f ) = x=0 ((x) f (x) - x( f (x))) d x. Show that I has a finite supremum on Ca and that there exists an f Ca at which that supremum is attained. Solution by Eugen J. Ionascu, Columbus State University, Columbus, GA. For every x [0, a] we let gx (u) = (x)u - x(u), defined for all u R. The derivative is gx (u) = (x) - x (u). By the mean value theorem, (x) = (x) - (0) = (x - 0) (cx ) for some cx between 0 and x. If x &gt; 0, then gx (u) = x( (cx ) - (u)). Because is strictly increasing, cx is uniquely determined, and gx attains its maximum at cx . If x = 0, then gx 0, and we simply define c0 = 0. This gives us a function x cx which we denote by f 0 . Clearly f 0 (x) = -1 (x)/x , 0, if 0 &lt; x a, if x = 0. [Monthly 117

280

c

THE MATHEMATICAL ASSOCIATION OF AMERICA

This function is continuous at every positive point x, since is continuous and strictly increasing. Also, because 0 &lt; cx &lt; x for x &gt; 0, this function is also continuous at 0. Thus, f 0 Ca . For all f Ca , we have

a a

I( f ) =

0

gx f (x) d x

0

gx f 0 (x) d x = I ( f 0 ).

This inequality answers both parts of the problem. Editorial comment. Richard Bagby noted that it is not necessary to explicitly assume the continuity of . If is differentiable everywhere, then has the intermediate value property by Darboux's theorem, and every monotonic function on R with the intermediate value property is continuous.

Points Generated by the Nine Points 11370 [2008, 568]. Proposed by Michael Goldenberg and Mark Kaplan, Baltimore Polytechnic Institute, Baltimore, MD. Let A0 , A1 , and A2 be the vertices of a nonequilateral triangle T . Let G and H be the centroid and orthocenter of T , respectively. Treating all indices modulo 3, let Bk be the midpoint of Ak-1 Ak+1 , let Ck be the foot of the altitude from Ak , and let Dk be the midpoint of Ak H . The nine-point circle of T is the circle through all Bk , Ck , and Dk . We now introduce nine more points, each obtained by intersecting a pair of lines. (The intersection is not claimed to occur between the two points specifying a line.) Let Pk be the intersection of Bk-1 Ck+1 and Bk+1 Ck-1 , Q k the intersection of Ck-1 Dk+1 and Ck+1 Dk-1 , and Rk the intersection of Ck-1 Ck+1 and Dk-1 Dk+1 . Let e be the line through {P0 , P1 , P2 }, and f be the line through {Q 0 , Q 1 , Q 2 }. (By Pascal's theorem, these triples of points are collinear.) Let g be the line through {R0 , R1 , R2 }; by Desargues' theorem, these points are also collinear. (a) Show that the line e is the Euler line of T . (b) Show that g coincides with f . (c) Show that f is perpendicular to e. (d) Show that the intersection S of e and f is the inverse of H with respect to the nine-point circle. Solution by the proposers. (a) Let k, m, n be 1, 2, 3 in some order. Applying Pappus's theorem to points Bm , Cm , An on line Ak An and to points Bn , Cn , Am on line Ak Am , we get that the three points Pk , G, and H , defined by Pk = Bm Cn Bn Cm , G = Am Bm An Bn , and H = Am Cm An Cn , are collinear. So all Pk lie on the Euler line G H . (b) Let N be the nine-point circle. Consider the cyclic quadrilateral Cm Cn Dm Dn . Because H = Cm Dm Cn Dn , Q k = Cm Dn Cn Dm , and Rk = Cm Cn Dm Dn , we conclude that points Q k and Rk are on the polar of H with respect to N (see Theorem 6.51, p. 145, in H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, Mathematical Association of America, Washington, DC 1967). So f and g coincide. (c) By the definition of polar, we have N H f or e f . (d) This also follows from the definition of polar. Editorial comment. Most solvers proceeded analytically. Some solvers simplified the algebra by using complex numbers or determinants. Some used Maple to help. March 2010]

For Grid Triangles, the Brocard Angle is Irrational in Degrees 11375 [2008, 568]. Proposed by Cezar Lupu, student, University of Bucharest, Bucharest, Romania. The first Brocard point of a triangle ABC is that interior point for which the angles BC, C A, and AB have the same radian measure. Let be that measure. Regarding the triangle as a figure in the Euclidean plane R2 , show that if the vertices belong to Z × Z, then / is irrational. Editorial comment. The claim follows from combining several well-known results. (a) cot = cot A + cot B + cot C = (a 2 + b2 + c2 )/4S 3, where S is the area of the triangle. The first equality is shown in [1]; see also [5] and [7]. The second is an easy consequence of the law of sines and the law of cosines. The inequality is due to Weitzenb¨ ck [2], also proved in [8]. o (b) Because the cotangent is decreasing on (0, /2), we conclude that /6. This is also deduced in [1] and [7]. (c) The squares of the sides (by the distance formula), the area S (by Pick's Theorem), and all six trigonometric functions of the angles (by various elementary trigonometric relationships) are rational because the vertices belong to Z × Z. (d) Every angle in (0, /2) that is a rational multiple of and has rational trigonometric functions is larger than /6 (using Lambert's theorem; see also [6]); so cannot be a rational multiple of .

Jerry Minkus showed that a similar result can be obtained for triangles whose vertices lie in the set of vertices of the unit triangular tiling of the plane, except that of course equilateral triangles (for which = /6) must be excluded. He also conjectured a generalization. Given a square-free positive integer d other than 3, let the lattice L d be defined by {h + k : h, k Z}, where = i d when d is congruent to 1 or 2 mod 4, and = (-1 + i d)/2 when d 3 (mod 4). The conjecture is that if the vertices of triangle ABC lie on L d , then the Brocard angle of triangle ABC is an irrational multiple of .

and limn Tn (a) = 2. Thus, Sn (a) converges if and only if (n - n {an})-1/2 does; that is, it converges when Rn (a) = n - n {an} has a positive limit, finite or infinite. If a is rational, then writing a = p/q with p and q relatively prime yields 1 - {an} 1/q, so Rn (a) n/q and Sn (a) converges. If a is irrational, then its continued fraction convergents pk /qk satisfy 0 &lt; a - 2 2 pk /qk &lt; 1/qk for even k, and 0 &lt; pk /qk - a &lt; 1/qk for odd k. Thus for even k, {qk a} &lt; 1/qk so that Rqk (a) qk - 1; on even k, this subsequence tends to infinity. For odd k, on the other hand, {qk a} &gt; 1 - 1/qk so that Rqk (a) 1; this subsequence remains bounded. Thus Rn (a) has neither a positive nor infinite limit, and therefore Sn (a) diverges. Editorial comment. Several solvers noted that Sn (a) is a Riemann sum for the expres a+1 sion a d x/ x - a, which evaluates to 2. Since the integral is improper, it need not equal the limit of its Riemann sums.

Angles of a Triangle 11385 [2008, 757]. Proposed by Jos´ Luis D´az-Barrero, Universidad Polit´ cnica de e i e Catalu~ a, Barcelona, Spain. Let 0 , 1 , and 2 be the radian measures of the angles of n an acute triangle, and for i 3 let i = i-3 . Show that

Calculation shows that the second derivative of (3 + 2 tan2 )1/4 is positive on (0l/2). Apply the AM­GM inequality to the first factor and Jensen's inequality on the second factor to obtain 1 3 i2 i+1 i+2 (3 + 2 tan2 i )1/4